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Recently I am reading the proof of the de Rham theorem. One step of the proof is to prove the singular cohomology $H^i(X,\mathbb{Z})$ is isomorphic to the sheaf cohomology $H^i(X,\mathbb{Z})$ of the constant sheaf $\mathbb{Z}$ on $X$. Here is a proof I found:proposition2.1

In step 3, the author gave an exact sequence as follows(here $\mathcal{V}$ is an open covering of $X$) : $0 \rightarrow ker \pi_v \rightarrow C_{sing}^*(X,\mathbb{Z}) \xrightarrow{\pi_v} C_{sing}^{\mathcal{V},*}(X,\mathbb{Z}) \rightarrow 0$. My question is: why is it surjective at the end of the sequence?

I know that there is a natural inclusion: $C_*^{\mathcal{V}}(X,\mathbb{Z}) \rightarrow C_*(X,\mathbb{Z})$, and by applying the hom functor $Hom(\bullet,\mathbb{Z})$ we can get a morphism $\pi_{\mathcal{V}}: C_{sing}^*(X,\mathbb{Z}) \rightarrow C_{sing}^{\mathcal{V},*}(X,\mathbb{Z})$, but I do not understand why it is surjective. I think it is just a simple question in algebraic topology, but I just started learning algebraic topology recently and may miss some easy facts. Thanks for any hints in advance.

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Let $i:C_* \rightarrow D_*$ be an inclusion of chain complexes where all groups are freely generated abelian groups by sets $SC_n$ and $SD_n$ and $i(SC_n) \subset SD_n$ as subsets of $C_n$ and $D_n$.

Then for a map $f:C_n \rightarrow A$ (this is an n-cochain). Define a map $f':D_n \rightarrow A$ where $f'$ is zero on elements of $SD_n$ that are not in $SC_n$ but equal to $f$ on elements of $SC_n$.

We can do this because of the category theoretical properties of the functor which takes a set to the group freely generated by it. This property says that any homomorphism of groups $g:FX \rightarrow A$ where $FX$ is the freely generated abelian group by the set $X$ is equivalent to defining a function $g:X \rightarrow A$ on the underlying sets.

Anyways, our construction of $f'$ make the surjectivity of $i^*:D^n \rightarrow C^n$ clear since $f' \circ i = f$ by definition.

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  • $\begingroup$ Oh right, thanks. $\endgroup$ Oct 17, 2019 at 5:18

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